Download SECTION 8-2 Mathematical Induction

560
8 Sequences and Series
61. Approximate e0.2 using the first five terms of the series.
Compare this approximation with your calculator evaluation of e0.2.
0.5
62. Approximate e
using the first five terms of the series.
Compare this approximation with your calculator evaluation of e0.5.
SECTION
8-2
ca
k
n
c
k1
a
n
64. Show that:
(a
k
k
k1
n
bk ) k1
n
a
k
k1
b
k
k1
Mathematical Induction
•
•
•
•
• Introduction
n
63. Show that:
Introduction
Mathematical Induction
Additional Examples of Mathematical Induction
Three Famous Problems
In common usage, the word “induction” means the generalization from particular
cases or facts. The ability to formulate general hypotheses from a limited number of
facts is a distinguishing characteristic of a creative mathematician. The creative
process does not stop here, however. These hypotheses must then be proved or disproved. In mathematics, a special method of proof called mathematical induction
ranks among the most important basic tools in a mathematician’s toolbox. In this section mathematical induction will be used to prove a variety of mathematical statements, some new and some that up to now we have just assumed to be true.
We illustrate the process of formulating hypotheses by an example. Suppose we
are interested in the sum of the first n consecutive odd integers, where n is a positive
integer. We begin by writing the sums for the first few values of n to see if we can
observe a pattern:
11
n 1
134
n 2
1359
n 3
1 3 5 7 16
n 4
1 3 5 7 9 25
n 5
Is there any pattern to the sums 1, 4, 9, 16, and 25? You no doubt observed that each
is a perfect square and, in fact, each is the square of the number of terms in the sum.
Thus, the following conjecture seems reasonable:
Conjecture Pn: For each positive integer n,
1 3 5 . . . (2n 1) n2
That is, the sum of the first n odd integers is n2 for each positive integer n.
So far ordinary induction has been used to generalize the pattern observed in the
first few cases listed above. But at this point conjecture Pn is simply that—a conjecture. How do we prove that Pn is a true statement? Continuing to list specific cases will
never provide a general proof—not in your lifetime or all your descendants’ lifetimes!
Mathematical induction is the tool we will use to establish the validity of conjecture Pn.
8-2
Mathematical Induction
561
Before discussing this method of proof, let’s consider another conjecture:
Conjecture Qn: For each positive integer n, the number n 2 n 41 is a prime
number.
TABLE 1
n
n2 n 41
1
41
Yes
2
43
Yes
3
47
Yes
4
53
Yes
5
61
Yes
Prime?
EXPLORE-DISCUSS 1
• Mathematical
Induction
Theorem 1
It is important to recognize that a conjecture can be proved false if it fails for
only one case. A single case or example for which a conjecture fails is called a counterexample. We check the conjecture for a few particular cases in Table 1. From the
table, it certainly appears that conjecture Qn has a good chance of being true. You
may want to check a few more cases. If you persist, you will find that conjecture Qn
is true for n up to 41. What happens at n 41?
412 41 41 412
which is not prime. Thus, since n 41 provides a counterexample, conjecture Qn is
false. Here we see the danger of generalizing without proof from a few special cases.
This example was discovered by Euler (1707–1783).
Prove that the following statement is false by finding a counterexample: If n 2,
then at least one-third of the positive integers less than or equal to n are prime.
We begin by stating the principle of mathematical induction, which forms the basis
for all our work in this section.
Principle of Mathematical Induction
Let Pn be a statement associated with each positive integer n, and suppose the
following conditions are satisfied:
1. P1 is true.
2. For any positive integer k, if Pk is true, then Pk1 is also true.
Then the statement Pn is true for all positive integers n.
Theorem 1 must be read very carefully. At first glance, it seems to say that if we
assume a statement is true, then it is true. But that is not the case at all. If the two
conditions in Theorem 1 are satisfied, then we can reason as follows:
P1 is true.
Condition 1
P2 is true, because P1 is true.
Condition 2
P3 is true, because P2 is true.
Condition 2
P4 is true, because P3 is true.
.
.
.
Condition 2
.
.
.
562
8 Sequences and Series
Condition 1: The first domino
can be pushed over.
(a)
Since this chain of implications never ends, we will eventually reach Pn for any positive integer n.
To help visualize this process, picture a row of dominoes that goes on forever
(see Fig. 1) and interpret the conditions in Theorem 1 as follows: Condition 1 says
that the first domino can be pushed over. Condition 2 says that if the kth domino falls,
then so does the (k 1)st domino. Together, these two conditions imply that all the
dominoes must fall.
Now, to illustrate the process of proof by mathematical induction, we return to
the conjecture Pn discussed earlier, which we restate below:
Pn:
1 3 5 . . . (2n 1) n2
n any positive integer
We already know that P1 is a true statement. In fact, we demonstrated that P1 through
P5 are all true by direct calculation. Thus, condition 1 in Theorem 1 is satisfied. To
show that condition 2 is satisfied, we assume that Pk is a true statement:
Condition 2: If the kth domino
falls, then so does the (k 1)st.
(b)
Pk : 1 3 5 . . . (2k 1) k 2
Now we must show that this assumption implies that Pk1 is also a true statement:
Pk1:
Conclusion: All the dominoes
will fall.
(c)
FIGURE 1 Interpreting mathematical induction.
1 3 5 . . . (2k 1) (2k 1) (k 1)2
Since we have assumed that Pk is true, we can perform operations on this equation.
Note that the left side of Pk1 is the left side of Pk plus (2k 1). So we start by
adding (2k 1) to both sides of Pk:
1 3 5 . . . (2k 1) k 2
1 3 5 . . . (2k 1) (2k 1) k 2 (2k 1)
Pk
Add 2k 1 to
both sides.
Factoring the right side of this equation, we have
1 3 5 . . . (2k 1) (2k 1) (k 1)2
Pk1
But this last equation is Pk1. Thus, we have started with Pk , the statement we assumed
true, and performed valid operations to produce Pk1, the statement we want to be
true. In other words, we have shown that if Pk is true, then Pk1 is also true. Since
both conditions in Theorem 1 are satisfied, Pn is true for all positive integers n.
• Additional
Examples of
Mathematical
Induction
EXAMPLE 1
Now we will consider some additional examples of proof by induction. The first is
another summation formula. Mathematical induction is the primary tool for proving
that formulas of this type are true.
Proving a Summation Formula
Prove that for all positive integers n
1 1 1 ...
1
2n 1
n
2 4 8
2
2n
8-2
Proof
563
State the conjecture:
Pn :
Part 1
Mathematical Induction
1 1 1 ...
1
2n 1
n
2 4 8
2
2n
Show that P1 is true.
P1:
1 21 1
2
21
1
2
Thus, P1 is true.
Part 2
Show that if Pk is true, then Pk1 is true. It is a good practice to always write out both
Pk and Pk1 at the beginning of any induction proof to see what is assumed and what
must be proved:
Pk :
Pk1:
1 1 1 ...
1
2k 1
k
2 4 8
2
2k
1 1 1 ...
1
1
2k1 1
k k1 2 4 8
2
2
2k1
We assume
Pk is true.
We must show that
Pk1 follows from Pk.
We start with the true statement Pk, add 1/2k1 to both sides, and simplify the right
side:
1 1 1 ...
1
2k 1
k
2 4 8
2
2k
Pk
1 1 1 ...
1
1
2k 1
1
k k1 k1
k
2 4 8
2
2
2
2
2k 1 2
1
k1
k
2
2 2
2k1 2 1
2k1
2k1 1
2k1
Thus,
1 1 1 ...
1
1
2k1 1
k k1 2 4 8
2
2
2k1
Pk1
and we have shown that if Pk is true, then Pk1 is true.
Conclusion
Both conditions in Theorem 1 are satisfied. Thus, Pn is true for all positive integers
n.
564
8 Sequences and Series
Matched Problem 1
Prove that for all positive integers n
123...n
n(n 1)
2
The next example provides a proof of a law of exponents that previously we had to
assume was true. First we redefine an for n a positive integer, using a recursion formula:
DEFINITION 1
Recursive Definition of a n
For n a positive integer
a1 a
a n 1 a na
EXAMPLE 2
n1
Proving a Law of Exponents
Prove that (xy)n x n y n for all positive integers n.
Proof
State the conjecture:
Pn:
Part 1
(xy)n xnyn
Show that P1 is true.
(xy)1 xy
x1 y1
Definition 1
Definition 1
Thus, P1 is true.
Part 2
Show that if Pk is true, then Pk1 is true.
Pk : (xy)k x k y k
k1
Pk1: (xy)
x
Assume Pk is true.
k1 k1
y
Show that Pk1 follows from Pk.
Here we start with the left side of Pk1 and use Pk to find the right side of Pk1:
(xy)k1 (xy)k(xy)1
Definition 1
x k y kxy
Use Pk: (xy)k x ky k
(x kx)(y k y)
Property of real numbers
x k1 y k1
Definition 1
8-2
Mathematical Induction
565
Thus, (xy)k1 x k1 y k1, and we have shown that if Pk is true, then Pk1 is true.
Conclusion
Matched Problem 2
Both conditions in Theorem 1 are satisfied. Thus, Pn is true for all positive integers
n.
Prove that (x/y) n x n/y n for all positive integers n.
Our last example deals with factors of integers. Before we start, recall that an
integer p is divisible by an integer q if p qr for some integer r.
EXAMPLE 3
Proving a Divisibility Property
Prove that 42n 1 is divisible by 5 for all positive integers n.
Proof
Use the definition of divisibility to state the conjecture as follows:
Pn :
Part 1
42n 1 5r
for some integer r
Show that P1 is true.
42 1 15 5 3
P1:
Thus, P1 is true.
Part 2
Show that if Pk is true, then Pk1 is true.
42k 1 5r
for some integer r
Assume Pk is true.
42(k1) 1 5s
for some integer s
Show that Pk1 must follow.
Pk :
Pk1:
As before, we start with the true statement Pk:
42k 1 5r
Pk
4 (4 1) 4 (5r)
Multiply both sides by 42.
42k2 16 80r
Simplify.
2
2k
2(k1)
4
2
1 80r 15
5(16r 3)
Add 15 to both sides.
Factor out 5.
Thus,
42(k1) 1 5s
Pk1
where s 16r 3 is an integer, and we have shown that if Pk is true, then Pk1
is true.
566
8 Sequences and Series
Conclusion
Matched Problem 3
Both conditions in Theorem 1 are satisfied. Thus, Pn is true for all positive
integers n.
Prove that 8n 1 is divisible by 7 for all positive integers n.
In some cases, a conjecture may be true only for n m, where m is a positive
integer, rather than for all n 0. For example, see Problems 49 and 50 in Exercise
8-2. The principle of mathematical induction can be extended to cover cases like this
as follows:
Theorem 2
Extended Principle of Mathematical Induction
Let m be a positive integer, let Pn be a statement associated with each integer
n m, and suppose the following conditions are satisfied:
1. Pm is true.
2. For any integer k m, if Pk is true, then Pk1 is also true.
Then the statement Pn is true for all integers n m.
• Three Famous
Problems
The problem of determining whether a certain statement about the positive integers
is true may be extremely difficult. Proofs may require remarkable insight and ingenuity and the development of techniques far more advanced than mathematical induction. Consider, for example, the famous problems of proving the following statements:
1. Lagrange’s Four Square Theorem, 1772: Each positive integer can be expressed
as the sum of four or fewer squares of positive integers.
2. Fermat’s Last Theorem, 1637: For n 2, x n y n z n does not have solutions
in the natural numbers.
3. Goldbach’s Conjecture, 1742: Every positive even integer greater than 2 is the
sum of two prime numbers.
The first statement was considered by the early Greeks and finally proved in 1772 by
Lagrange. Fermat’s last theorem, defying the best mathematical minds for over 350 years,
finally succumbed to a 200-page proof by Prof. Andrew Wiles of Princeton University
in 1993. To this date no one has been able to prove or disprove Goldbach’s conjecture.
EXPLORE-DISCUSS 2
(A) Explain the difference between a theorem and a conjecture.
(B) Why is “Fermat’s last theorem” a misnomer? Suggest more accurate names
for the result.
8-2
Mathematical Induction
567
Answers to Matched Problems
1. Sketch of proof. State the conjecture: Pn: 1 2 3 . . . n n(n 1)
2
1(1 1)
. P1 is true.
2
Part 2. Show that if Pk is true, then Pk1 is true.
Part 1. 1 123...k
1 2 3 . . . k (k 1) Conclusion: Pn is true.
xy yx
n
2. Sketch of proof. State the conjecture: Pn:
xy xy xy . P
1
Part 1.
k(k 1)
2
Pk
k(k 1)
(k 1)
2
(k 1)(k 2)
2
Pk1
n
n
1
1
1
is true.
Part 2. Show that if Pk is true, then Pk1 is true.
xy k1
xy xy y xy y xy xy
k
xk
xk
k1
k
k
k1
Conclusion: Pn is true.
3. Sketch of proof. State the conjecture: Pn: 8n 1 7r
Part 1. 81 1 7 7 1. P1 is true.
Part 2. Show that if Pk is true, then Pk1 is true.
for some integer r
8k 1 7r
Pk
8(8k 1) 8(7r)
8k1 1 56r 7 7(8r 1) 7s
Pk1
Conclusion: Pn is true.
EXERCISE
8-2
A
10. Pn : 4 n 1 is divisible by 3
In Problems 1–4, find the first positive integer n that causes
the statement to fail.
Write Pk and Pk1 for Pn as indicated in Problems 11–16.
1. 3n 4n 5n
2. n2 3n 100
3. 17n 1 is divisible by 2n
4. n2 5n 6
Verify each statement Pn in Problems 5–10 for n 1, 2,
and 3.
5. Pn : 2 6 10 . . . (4n 2) 2n 2
6. Pn : 4 8 12 . . . 4n 2n(n 1)
7. Pn : a a a
5
n
5n
9. Pn : 9 1 is divisible by 4
n
8. Pn : (a ) a
5 n
5n
11. Pn in Problem 5
12. Pn in Problem 6
13. Pn in Problem 7
14. Pn in Problem 8
15. Pn in Problem 9
16. Pn in Problem 10
In Problems 17–22, use mathematical induction to prove
that each Pn holds for all positive integers n.
17. Pn in Problem 5
18. Pn in Problem 6
19. Pn in Problem 7
20. Pn in Problem 8
21. Pn in Problem 9
22. Pn in Problem 10
568
8 Sequences and Series
41. 13 23 33 . . . n3 (1 2 3 . . . n)2
[Hint: See Matched Problem 1 following Example 1.]
B
In Problems 23–26, prove the statement is false by finding a
counterexample.
42.
23. If n 2, then any polynomial of degree n has at least one
real zero.
24. Any positive integer n 7 can be written as the sum of
three or fewer squares of positive integers.
25. If n is a positive integer, then there is at least one prime
number p such that n p n 6.
26. If a, b, c, d are positive integers such that a2 b2 c2 d 2,
then a c or a d.
In Problems 27–42, use mathematical induction to prove
each proposition for all positive integers n, unless restricted
otherwise.
27. 2 22 23 . . . 2n 2n1 2
28.
1
1 1 1 ...
1
n1
2 4 8
2
2
n
n(n 1)(2n 1)
31. 1 2 3 . . . n2 6
2
n(n 1) an
an3; n 3
a3
34.
n(n 1)(n 2)
3
n
44.
1
1
1
1
...
12 23 34
n(n 1)
In Problems 47–50, prove the statement is true for all integers n as specified.
47. a 1 ⇒ a n 1; n N
48. 0 a 1 ⇒ 0 a n 1; n N
50. 2n n2; n 5
51. Prove or disprove the generalization of the following two
facts:
33 3 2 4 2 5 2
1
a5
;n5
a n a n 5
35. a a a ; m, n N
[Hint: Choose m as an arbitrary element of N, and then use
induction on n.]
m
43. 2 4 6 . . . 2n
49. n2 2n; n 3
32. 1 2 2 3 3 4 . . .
33.
In Problems 43–46, suggest a formula for each expression,
and prove your hypothesis using mathematical induction,
n N.
46. The number of diagonals in a polygon with n sides
30. 1 8 16 . . . 8(n 1) (2n 1)2; n 1
2
C
45. The number of lines determined by n points in a plane, no
three of which are collinear
29. 12 32 52 . . . (2n 1)2 13 (4n3 n)
2
1
1
1
...
123 234 345
1
n(n 3)
n(n 1)(n 2) 4(n 1)(n 2)
33 43 53 63
mn
36. (a n)m a mn ; m, n N
37. x n 1 is divisible by x 1; x 1
[Hint: Divisible means that x n 1 (x 1)Q(x) for some
polynomial Q(x).]
52. Prove or disprove: n2 21n 1 is a prime number for all
natural numbers n.
If {an} and {bn} are two sequences, we write {an} {bn} if
and only if an bn , n N. In Problems 53–56, use mathematical induction to show that {an} {bn}.
53. a1 1, an an1 2; bn 2n 1
38. x y is divisible by x y; x y
54. a1 2, an an1 2; bn 2n
39. x 1 is divisible by x 1; x 1
55. a1 2, an 22an1; bn 22n1
40. x 2n 1 is divisible by x 1; x 1
56. a1 2, an 3an1; bn 2 3n1
n
n
2n
SECTION
8-3
Arithmetic and Geometric Sequences
•
•
•
•
•
Arithmetic and Geometric Sequences
nth-Term Formulas
Sum Formulas for Finite Arithmetic Series
Sum Formulas for Finite Geometric Series
Sum Formula for Infinite Geometric Series